# Upper bound for a interesting function.

Prove $$\frac{1}{2}^x+\frac{1}{2}^\frac {1}{x}\leq 1$$, where $x$ is a positive real number. This problem is from my friend. Here is my approach. It is sufficient to show that $$0\leq2^x-1-2^{x-\frac {1}{x}}$$. To find minimum of right side, I differentiate the right fuction, but I cannot find all zeros of $2^t-1-t^2$, where $t=\frac{1}{x}$. Do you have any idea to find upperbound?

• They meet at t=0,1 and one more point between 4 and 5. – user148928 Jun 9 '15 at 2:58
• The problem is it takes a minimum at a point between 4 and 5. – user148928 Jun 9 '15 at 3:19
• You need to worry only about $x\in (0,1)$, the LHS is invariant to $t \mapsto \frac1t$. – Macavity Jun 9 '15 at 3:21
• take $f(t)=ln2*t-ln(1+t^2)$,prove on(0,1) there is only one max positive point, so two ends will be min. , – chenbai Jun 9 '15 at 3:29
• @Macavity Thank you for your comment. – user148928 Jun 9 '15 at 4:12

Considering the function $$f(x)=1-2^{-1/x}-2^{-x}$$ As Macavity commented, $f(\frac1x)=f(x)$ and you only need to consider $x\in (0,1)$. Look at the derivatives $$f'(x)=2^{-x} \log (2)-\frac{2^{-1/x} \log (2)}{x^2}$$ $$f''(x)=-\frac{2^{-1/x} \log ^2(2)}{x^4}+\frac{2^{1-\frac{1}{x}} \log (2)}{x^3}-2^{-x} \log ^2(2)$$ Using limits for the lower bound $x=0$, you can show that $$f(0)=0 \\\ f'(0)=\log (2) \\\ f''(0)=-\log ^2(2)$$ Similarly $$f(1)=0 \\\ f'(1)=0 \\\ f''(1)=\log (2)-\log ^2(2)$$ Looking at the first derivative, it cancels close to $x=0.2$; using Newton method starting at this point, the successive iterates are $0.214110$, $0.215101$, $0.215106$ which is the solution for six significant figures. At this point, the second derivative is $-3.80712$ which confirms that the point is a maximum.
So, for the considered interval, $f(x)\geq 0$ and it reaches a maximum value $\approx 0.0986559$ at $x\approx0.215106$.